/*
To the Max
Time Limit: 1 Second      Memory Limit: 32768 KB

Problem

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 1 or greater located within the whole array. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the sub-rectangle with the largest sum is referred to as the maximal sub-rectangle.

As an example, the maximal sub-rectangle of the array:

0 -2 -7 0
9 2 -6 2
-4 1 -4 1
-1 8 0 -2

is in the lower left corner:

9 2
-4 1
-1 8

and has a sum of 15.

The input consists of an N x N array of integers. The input begins with a single positive integer N on a line by itself, indicating the size of the square two-dimensional array. This is followed by N 2 integers separated by whitespace (spaces and newlines). These are the N 2 integers of the array, presented in row-major order. That is, all numbers in the first row, left to right, then all numbers in the second row, left to right, etc. N may be as large as 100. The numbers in the array will be in the range [-127,127].


Output

Output the sum of the maximal sub-rectangle.


Example

Input

4
0 -2 -7 0 9 2 -6 2
-4 1 -4 1 -1
8 0 -2

Output

15 
*/

#include <iostream>
using namespace std;


int N=0;
int **M;
int ***Row;
int ***Col;

int calc(int lr,int lc, int rr,int rc) {
	if (lr>=rr || lc>=rc) return 0;

   int ret = 0;	
   for (int i=lr;i<=rr;i++) 
	   ret += Row[i][lc][rc];

   	return ret;
}


int main(){
	int max=-127;
	int t,count=0;

	cin>>N;
	M = new int* [N];
	Row = new int** [N];
	Col = new int** [N];
	for (int i=0;i<N;i++) {
		M[i]=new int[N];
		Row[i]=new int* [N];
	}

	for (int i=0;i<N;i++)
		for (int j=0;j<N;j++) 
			Row[i][j]=new int[N];

	while (cin>>t) {
		if (max<t) max = t;
		M[count/N][count%N] = t;
		++ count;
	}

	for (int i=0;i<N;i++) 
		for (int j=0;j<N;j++)
				Row[i][j][j] = M[i][j];

	for (int i=0;i<N;i++) 
		for (int j=0;j<N;j++) 
			for (int k=j+1;j<N;j++) 
				Row[i][j][k] = Row[i][j][k-1]+M[i][k];

	for (int i=0;i<N;i++) 
		for (int j=0;j<N;j++) 
			for (int k=i+1;k<N;k++) 
				for (int x=j+1;x<N;x++) {
					t=calc(i,j,k,x);
					if (t>max) 
						max = t;
				}
	
	cout<<max<<endl;
}
